58 BULLETIN OF THE 



whence 



26_l_^2a_l^l 



a' h ^-^ a c ^ ^ 



Again 



^- = cos i B == cos (45°_ i A) = ^iAdb^S^M J 



whence 



sin^A = — I cos^A = 



/3 ^ 



and since sin^^A + cos' -^ A = 1, 



a / a 



or 



2a' 2x/2a& , 26' ., ,., 



If now we eliminate h between Eqs. (3) and (4) we have are 

 equation from which a may be found. 



From (4) we find, h = — ^ \a ±.V s^ a^ \ which substitut- 



/3 \/2 t. J 



ed in (3) gives after some reduction 



2 a* — 13^ dr2ma ^/^ga qJi 



<* a±:>//3' — a^ 



where wi = - v/2. This equation finally reduces to 

 tt 



(a*_a|3>/2+|30a«'— (3a^ — 3v/2a/3 + 2i3^)^a«+ (3a' — 

 2v/2ttp)— a' — — = 0. (5) 



THIRD SOLUTION. 



Revolve the triangles B E and D A about B and A O 



respectively so that E falls upon E' and D upon D', then 



E B = E' O B = E'OD'= D'OA = AOD = 45°, 



and consequently BOD' and A E' are right-angled triangles : 



hence 



2^ = tan i A, or -^ = 1 + tan i A ; 

 OA ^ OA ^ 2 ' 



whence a = O A (1 + tan \ A), (6) 



and similarly ,6 = O B (1 + tan ^ B). (7) 



